Atomic and Molecular Structure By: Kevin “Diminutive Vocabulary” Mospan and Liz “Bottle Cap Tea Cup” Schultz.

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Presentation transcript:

Atomic and Molecular Structure By: Kevin “Diminutive Vocabulary” Mospan and Liz “Bottle Cap Tea Cup” Schultz

Lewis Structures: Do you know how to draw them? Molecular Shapes: That fun process when you try to represent 3D things on 2D paper Formal charge: Remember that? Closer to 0 is favorable, and means that you’re Lewis structure is correct…probably Bond polarity: The difference in the electronegativity needs to be between 0.4 and 1.7 for the bond to be polar Molecular polarity: symmetrical means things cancel and are therefore not polar All of these are things you can practice on our quiz! How exciting!

Electromagnetic radiation .  = c = x 10 8 m/s Wavelength (  )– the distance between successive peaks or troughs Frequency – the number of complete waves or cycles passing thru a point in a certain amount of time Amplitude – the maximum height of the wave (p.295)

Relating Wave Properties and Energy Useful equation: E = h.  Energy of a photon is inversely proportional to the wavelength. i.e. ↑  means ↓ Energy Energy of a photon is directly proportional to the frequency. i.e. ↑  means ↑ Energy There are 6.02 x photons of energy per mole of light Avogadro's Number strikes again!

Planck’s constant is h ≈ x J. s Useful equation: E = h.  Photoelectric effect: Phenomenon explained by Einstein using Planck’s ideas Occurs when light strikes the surface of a metal, and electrons are ejected. Another useful equation: E = h. c / 

Electrons lie in circular orbits around molecule Radius of the circular orbits increases as n increases An atom with its electrons in the lowest possible energy levels is said to be in its “ground state” When an electron occupies an orbit greater than the lowest possible energy level, it is said to be in an “excited state” Somewhat useful equation: ΔE = -Rhc 1 1 Rhc ≈ 1312 kJ/mol - nf2nf2 () ni2ni2

Not with swords. Or guns. Wave/Particle duality applies what Einstein and Planck said about light photons to massless electrons. Electrons also exhibit both wave and particle properties. This is why we can use those “useful” light equations on the previous slides for electrons as well.

Actually, it’s DeBroglie and he had some valid things to say. For example: He utilized the idea of wave/particle duality to create yet another “useful” equation:  = h m.  Just make sure m (mass) is in kg

Value of l Corresponding subshell label 0s 1p 2d 3f There are 4 Quantum Numbers: n, l, m l, m s n is the principle quantum number it can be any integer from 1 to  l is the angular momentum quantum number. It characterizes the subshell of an orbital, like in the table It starts at 0 and goes to n-1 m l is the magnetic quantum number 0,  1,  2, …,  l Number of values for it in a given subshell is 2 l + 1 M s is the electron spin magnetic quantum #. It’s only ½ or -1/2

Orbitals are the places where electrons like to kick it. The lowest energy orbital is s, which has 1 subshell. Next is p, with 3 subshells, then d with 5 subshells, followed by f with 7 subshells. There are subshells after these, but they don’t really matter for the AP test because there are no known elements to fill them

Here’s the easiest way to figure out the electron configuration : All you have to do is alter the Periodic Table by putting the Lanthanides in the proper places as shown by the picture. Then you find your element. That’s basically it.

After you’ve figured out what the electron configuration is by using the period table, you just fill in the little boxes until you’ve filled everything up to and including the configuration you found. n is the number preceding the orbital for the selected atom l is still like it was in the table m l goes from the lowest number to the highest, left to right. M s has the +1/2 pointing up and the -1/2 pointing down The Pauli exclusion principle says that no 2 electrons can have the same set of all 4 quantum numbers 1s 2s 2p 3s 3p4s 3d 4p Quantum numbers for electron configurations

There are 3 major periodic trends 1.Radius 2.Electronegativity 3.Ionization Energy    

Radius As one goes from right to left, the number of protons decreases, so there are less intramolecular attractions causing the orbitals to expand As one goes from top to bottom, there are new orbitals which take up more space and increase the radius Cations are smaller than the original atom because it loses electrons which causes the orbitals to empty or constrict Anions are larger than the original atom because there is less force of proton per electron, so orbital can spread out farther

Electronegativity

Ionization Energy

This theory mainly deals with hybridization. Here’s a helpful table for hybridization if you can see it It’s also on your summary cheat sheet.

Valence bond theory says that there are 2 types of bonds: Sigma (σ) bonds and Pi (Π) bonds Sigma bonds occur when hybridized orbitals directly overlap. Pi bonds form from the unhybridized orbitals left after hybridization. In order for a pi bond to form, there must first be a sigma bond, because sigma bonds are much stronger.

Probably the most important part about molecular bond theory is that there are two kinds of orbitals, bonding orbitals and antibonding orbitals. In these orbitals, the antibonding orbitals are of much higher energy than the bonding orbitals. A group of atoms is “stabilized” and form chemical bonds when they are in the lower energy, bonding orbitals

The last principle of molecular bond theory is that electrons are assigned to orbitals of successively higher energy. This means that if you have a chart like that to the right, you start in the bottom and fill electrons in the spaces going up. Also, it’s important to remember that you fill in this chart with all of a molecule’s electrons, not just the valence electrons.

This nifty chart also helps illustrate how a filled bonding orbital and a filled antibonding orbital cancel each other out. If both orbitals are filled with electrons, there is no net bond for this reason. Lastly, bond order = (1/2)(# of electrons in bonding orbitals - # of electrons in antibonding orbitals)